CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2008

  • question_answer
    Eight drops of a liquid of density p and each of radius a are falling through air with a constant velocity\[3.75\,cm{{s}^{-1}}\]. When the eight drops coalesce to form a single drop the terminal velocity of the new drop will be

    A)  \[1.5\times {{10}^{-2}}m{{s}^{-1}}\]  

    B)         \[2.4\times {{10}^{-2}}m{{s}^{-1}}\]

    C)         \[0.75\times {{10}^{-2}}m{{s}^{-1}}\]

    D)         \[25\times {{10}^{-2}}m{{s}^{-1}}\]

    E)  \[15\times {{10}^{-2}}m{{s}^{-1}}\]

    Correct Answer: E

    Solution :

    Terminal velocity of a single drop, \[v=3.75\text{ }cm{{s}^{-1}}=3.75\times {{10}^{-2}}m{{s}^{-1}}\] Terminal velocity of the big drop \[V={{n}^{2/3}}v={{(8)}^{2/3}}\times 3.75\times {{10}^{-2}}\]                 \[=4\times 3.75\times {{10}^{-2}}\]                 \[=15\times {{10}^{-2}}m{{s}^{-1}}\]


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